Suppose $\tan \alpha = \dfrac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Prove that the number $\tan \beta$ for which $\tan {2 \beta} = \tan {3 \alpha}$ is rational only when $p^2 + q^2$ is the square of an integer.
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This is just some calculations!
we can easily show that $ \tan3\alpha=\frac{3\tan\alpha-\tan^{3}\alpha}{1-3\tan^{2}\alpha}=\frac{3\times\frac{p}{q}-\frac{p^{3}}{q^{3}}}{1-3\times\frac{p^{2}}{q^{2}}}=\frac{p\left(3q^{2}-p^{2}\right)}{q\left(q^{2}-3p^{2}\right)}$
now let $ \frac{1}{A}=\frac{p\left(3q^{2}-p^{2}\right)}{q\left(q^{2}-3p^{2}\right)}$
and as said in the problem:
$ \tan2\beta=\frac{2\tan\beta}{1-tan^{2}\beta}=\tan3\alpha=\frac{1}{A}\Longrightarrow\tan^{2}\beta+2A\tan\beta-1=0\Longrightarrow\tan\beta=-A+(\text{or }-)\sqrt{A^{2}+1}$
so $ \tan\beta$ is rational if and only if $ A^{2}+1$ is perfect square!
$ A^{2}+1=\frac{q^{2}\left(q^{2}-3p^{2}\right)^{2}+p^{2}\left(3q^{2}-p^{2}\right)^{2}}{p^{2}\left(3q^{2}-p^{2}\right)^{2}}$
so its sufficient that $ q^{2}\left(q^{2}-3p^{2}\right)^{2}+p^{2}\left(3q^{2}-p^{2}\right)^{2}=\left(p^{2}+q^{2}\right)^{3}$ is perfect square but this is only when $ p^{2}+q^{2}$ is a perfect square and we're done.